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14.Probability
easy
The probability of choosing at random a number that is divisible by $6$ or $8$ from among $1$ to $90$ is equal to
A
$\frac{1}{6}$
B
$\frac{1}{{30}}$
C
$\frac{{11}}{{80}}$
D
$\frac{{23}}{{90}}$
Solution
(d) We have, the number of divisible by $6$ in the series form $1$ to $90$ is $15$.
$\therefore $ Number divisible by $8$ in the series from $1$ to $90$ is equal to $11$ and the number divisible by both $6$ and $8$ in series from $1$ to $90 = 3.$
$\therefore $ Probability of choosing the number divisible by $6$ or $8$ is
$P = \frac{{15 + 11 – 3}}{{90}}$ or $P = \frac{{23}}{{90}}$.
Standard 11
Mathematics
